Can We Solve A Nonstandard Equation? 😮
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- Опубліковано 30 чер 2024
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Before even viewing, I smell the obnoxious odor of Lambert's W function.
🙃😜
So did I 😒
Two is clearly one solution. About two seconds to notice that.
After we guess x=2, we have to show that this solution is unique. Consider the function f(x)=x5^x. Note that for non positive x, f(x) is also non positive. and for positive x, f(x) is strictly increasing (no need to differentiate to see that). so there is a unique solution to our equation.
That's what I thought of instantly.
x5^x = (2)5²
xe^(xln5) = (2)5²
(xln5)e^(xln5) = (2ln5)5²
(xln5)e^(xln5) = (ln5²)5²
(xln5)e^(xln5) = (ln5²)(e^ln5²)
xln5 = ln5² = 2ln5
*x = 2*
Why use Lambert’s W Function when you can easily express that 50 = 2 * 25 = 2 * 5^2, meaning x = 2. It’s a similar process relative to using the W(), but much faster 😑
Because we’ve to confirm that 2 is the only solution. Sometimes we may get a complex solution via Lambert’s function. Taking derivative of x•5^x is to confirm 2 is the only real solution. This step is redundant if we’ve used Lambert’s function.
The prime factorization of 50 is 2*5^2. By 1:1 correspondence x = 2. It’s not “guess and check”.
Yes it is. Your logic is only valid if we were looking for integers x.
I did this. The following step is to know if there are more solutions; which is what Syber do at the end of the video. 👌
I got x=2 right away, but I could sense Lambert's W function being used just by looking at the original equation.
First impression: x = 2 is an obvious solution.
If we sketch the graphs y = 5^x and y = 50/x it would seem there can only be the one intersection, therefore x = 2 is the only solution. Now I'll watch the video. I am sure there will be some subtlety that has escaped me. This is a wonderful channel.
It's clever to set y = 5^x, so y=50/x. The exponential function is increasing everywhere, and the rational function is decreasing everywhere, and both functions are continuous for x>0, so they have at most one point of intersection
good point!
"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!
🫖
The function is strictly increasing and continuous, so it's invertible, so the obvious real solution is unique in the reals. But if you want complex solutions also, then lambert W is obviously the correct approach
Use Desmos and graph y= x*5^x - 50.
x=W(50ln5)/ln5
x×5^x=50 --> 5×5^x=2×5²
Thus x=2
I think we can also find the solution factoring the right side od the equation and easily proove the funtion is monotonic finding its derivative as always positive, what causes only one possibile solution found by factoring😁
8:30 Go through all these * hoops * . Not "loopholes".
I don't quite get why you try to do this dance inside the W function. You can start from the start by saying 50 = 2*25, and recognize that 25 = 5^2. So you rewrite the equation as x * 5^x = 2 * 5^2
Since the function on the left side is trvially increasing, there's only a single solution for x > 0, you can immediately see that x = 2 is the solution.
2 * 5^2 = 50 ho-hum
x = 2
🤑
😜
Seems like this is overcomplicating it to me..
"t"(tea) or coffee. Ha ha I love it! Better than to "b" or not to be!
☕️